Supporting Nested Resources in MrsP
Lemma 6 proof
Jorge Garrido1 , Shuai Zao2 , Alan Burns2 , and Andy Wellings2
1
Sistemas de Tiempo Real e Ingeniera de Servicios Telemticos (STRAST)
Universidad Politcnica de Madrid (UPM)
2
Department of Computer Science, University of York
Lemma 6.PThe cost of each individual access (e0 ) to a resource rj is bounded
by e0j = cj + rk ∈U(rj ) nkj ek .
Proof. As a consequence of Rule 1, there is at least one terminal resource
rt in the system not accessing any inner resource, i.e. U(rt ) = ∅. For such a
resource, its individual access cost is:
e0t = ct
From Lemma 5, if we simplify the queue of a resource as q j = |V (rj )| +
|map(G(rj ))| then the total access cost to et is:
et = q t ct ⇒ et = q t e0t
For the set of resources accessing the terminal resource, V(rt ), the individual
access cost can be expressed as the execution time of the resource plus the access
cost to its inner resource rt as:
e0t+1 = ct+1 + ntt+1 (q t ∗ ct )
then substituting et :
e0t+1 = ct+1 + ntt+1 et
And its total access cost can be again expressed as the queue for accessing
the resource times the cost of accessing it.
et+1 = q t+1 (ct+1 + ntt+1 et )
For the 2nd iteration of t outer resources:
t+1 t+1
e0t+2 = ct+2 + nt+1
(c
+ ntt+1 (q t ∗ ct )))
t+2 (q
t+1 t+1
e0t+2 = ct+2 + nt+1
(c
+ ntt+1 et ))
t+2 (q
t+1
e0t+2 = ct+2 + nt+1
)
t+2 (e
Then, for the level k of nesting:
t+1 t+1
e0k = ck + nk−1
(q k−1 (ck−1 + nk−2
(c
+ ntt+1 (q t ∗ ct )))))
k
k−1 (...q
t+1 t+1
e0k = ck + nk−1
(q k−1 (ck−1 + nk−2
(c
+ ntt+1 (et )))))
k
k−1 (...q
t+1
e0k = ck + nk−1
(q k−1 (ck−1 + nk−2
)))
k
k−1 (...e
...
k+2
e0k = ck + nk−1
(q k−1 (ck−1 + nk−2
))
k
k−1 e
e0k = ck + nk−1
ek−1
k
This can be directly applied to resources sequentially requiring more than
one different independent inner resources:
X
e0k = ck +
nk−1
ek−1
k
r k−1 ∈U(r k )